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            <h1 style="display: none">report-P3402 可持久化并查集</h1>
            
              <p class="note note-info">
                
                  本文最后更新于：1 年前
                
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            <div class="markdown-body" id="post-body">
              <p><strong><a target="_blank" rel="noopener" href="https://www.luogu.com.cn/problem/P3402">传送门</a></strong></p>
<h3 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述"></a>题目描述</h3><p>给定 $n$ 个集合，第 $i$ 个集合内初始状态下只有一个数，为 $i$。</p>
<p>有 $m$ 次操作。操作分为 $3$ 种：</p>
<ul>
<li><code>1 a b</code> 合并 $a$,$b$ 所在集合；</li>
<li><code>2 k</code> 回到第 $k$ 次操作（执行三种操作中的任意一种都记为一次操作）之后的状态；</li>
<li><code>3 a b</code> 询问 $a$,$b$ 是否属于同一集合，如果是则输出 $1$ ，否则输出 $0$。</li>
</ul>
<h3 id="输入格式"><a href="#输入格式" class="headerlink" title="输入格式"></a>输入格式</h3><p>第一行两个整数，$n$,$m$。</p>
<p>接下来 $m$ 行，每行先输入一个数 $opt$。若 $opt=2$ 则再输入一个整数 $k$，否则再输入两个整数 $a$,$b$，描述一次操作。</p>
<h3 id="样例"><a href="#样例" class="headerlink" title="样例"></a>样例</h3><div class="hljs"><pre><code class="hljs basic"><span class="hljs-symbol">5 </span><span class="hljs-number">6</span>
<span class="hljs-symbol">1 </span><span class="hljs-number">1</span> <span class="hljs-number">2</span>
<span class="hljs-symbol">3 </span><span class="hljs-number">1</span> <span class="hljs-number">2</span>
<span class="hljs-symbol">2 </span><span class="hljs-number">0</span>
<span class="hljs-symbol">3 </span><span class="hljs-number">1</span> <span class="hljs-number">2</span>
<span class="hljs-symbol">2 </span><span class="hljs-number">1</span>
<span class="hljs-symbol">3 </span><span class="hljs-number">1</span> <span class="hljs-number">2</span></code></pre></div>
<div class="hljs"><pre><code class="hljs angelscript"><span class="hljs-number">1</span>
<span class="hljs-number">0</span>
<span class="hljs-number">1</span></code></pre></div>
<hr>
<h3 id="思路分析"><a href="#思路分析" class="headerlink" title="思路分析"></a>思路分析</h3><p>这道题看似是个新的可持久化数据结构实际上还是一个主席树，只不过用来维护并查集而已。</p>
<p>看到题目里的可持久化，版本回跳等操作，我们自然而然的能联想到主席树。</p>
<p>关于主席树，可以参考<strong><a target="_blank" rel="noopener" href="https://weepingdemon.gitee.io/blog/2020/08/22/algHJTtree/">这篇文章</a></strong></p>
<hr>
<h4 id="实现可持久化"><a href="#实现可持久化" class="headerlink" title="实现可持久化"></a>实现可持久化</h4><p>主席树能实现什么，维护一棵能够保存旧版本的树，并且每次操作规模都很小，正好符合这里的需求。而该题目需要维护的是各个版本的并查集关系。</p>
<p>我们来联想一下并查集，不加路径压缩的并查集就是通过实现给每一个节点赋值<code>father</code>来实现的，也就是说，只要我们用主席树来维护这个<code>father</code>，我们就能初步<strong>实现</strong>可持久化并查集，只不过需要一些优化了。</p>
<hr>
<h4 id="提高效率"><a href="#提高效率" class="headerlink" title="提高效率"></a>提高效率</h4><p>我们考虑最需要优化的地方，并查集的效率主要都靠路径压缩来增加，然而现在由于是在主席树里实现的，我们每次只能操作比较少的点，所以路径压缩显然需要被抛弃。于是复杂度就上来了。</p>
<p>但还是有办法降低期望复杂度的，那就是启发式合并。我们考虑一个并查集找老大的时候的最坏情况，那就是所有节点都倚靠<code>father</code>的关系形成一条有向的链，这个时候显然是<code>get_father()</code>最慢的时候。所以我们要尽可能避免这种情况。</p>
<p>考虑我们现在有两个并查集的老大 $a$ 和 $b$ ，两个集合借由<code>father</code>的关系分别能够形成一棵树，且 $a$ 和 $b$ 的<code>father</code>都是自己，假设以每一个老大 $x$ 为根的树中深度最大的叶子节点的深度为<code>deep[x]</code>，并且我们假设<code>deep[a]</code> &lt;= <code>deep[b]</code>，那我们令 $a$ 的<code>father</code>为 $b$ 的<code>father</code>也就是 $b$ 自己，此时的<code>deep[b]</code>就等于<code>deep[b] + (deep[b] == deep[a])</code>，很容易理解，当 $a$ 和 $b$ 的最大深度相同的时候，把 $a$ 变成到 $b$ 的儿子总会使 $b$ 的最大深度增加 $1$ 。当然，如果<code>deep[a]</code>严格小于<code>deep[b]</code>就不会出现这么多问题。</p>
<p>所以其实整题的难度都聚集在主席树的构造和启发式合并上。</p>
<hr>
<h4 id="具体实现"><a href="#具体实现" class="headerlink" title="具体实现"></a>具体实现</h4><p>这道题有这么几个注意点：</p>
<ul>
<li>并查集不使用路径压缩   -&gt;   <strong>这就决定了我们要按秩合并</strong></li>
<li>主席树维护每一个元素各自的信息，实际上只用了树形数据结构的叶子节点，其他节点帮助实现二分查找</li>
</ul>
<p>我们先给出主席树需要维护的东西，以节点<code>cur</code>为例：</p>
<ul>
<li><code>cur-&gt;l</code>和<code>cur-&gt;r</code>：维护边界</li>
<li><code>cur-&gt;ls</code>和<code>cur-&gt;rs</code>：左右儿子</li>
<li><code>cur-&gt;deep</code>：节点深度，仅在叶子节点起作用，当且一个节点作为其所在集合的老大时表示集合的最大深度(即从集合内任意一个节点递归寻找到老大所需要的最大步数)</li>
<li><code>cur-&gt;fa</code>：节点在并查集中的父亲，仅在叶子节点起作用，用于递归寻找集合老大</li>
<li><code>cur-&gt;num</code>：节点所代表的数值，其实不起作用，其值等于叶子节点的<code>cur-&gt;l</code>或者<code>cur-&gt;r</code>（在叶子节点两者等价）</li>
</ul>
<p>首先，根据题目要求，维护版本的操作通过主席树实现，版本的回跳等都一样。</p>
<hr>
<h5 id="建树"><a href="#建树" class="headerlink" title="建树"></a>建树</h5><p>就是正常的主席树建树</p>
<div class="hljs"><pre><code class="hljs cpp"><span class="hljs-function"><span class="hljs-keyword">inline</span> <span class="hljs-keyword">void</span> <span class="hljs-title">build</span><span class="hljs-params">(node * cur,<span class="hljs-keyword">int</span> l,<span class="hljs-keyword">int</span> r)</span></span>&#123;
    cur-&gt;l = l , cur-&gt;r = r;
    <span class="hljs-keyword">if</span>(l == r)&#123;
        cur-&gt;fa = cur-&gt;num = l , cur-&gt;deep = <span class="hljs-number">1</span>;
        <span class="hljs-keyword">return</span>;
    &#125;
    <span class="hljs-keyword">int</span> mid = (l+r)&gt;&gt;<span class="hljs-number">1</span>;
    cur-&gt;ls = create() , cur-&gt;rs = create();
    build(cur-&gt;ls,l,mid) , build(cur-&gt;rs,mid+<span class="hljs-number">1</span>,r);
    <span class="hljs-keyword">return</span>;
&#125;</code></pre></div>
<hr>
<h5 id="找老大"><a href="#找老大" class="headerlink" title="找老大"></a>找老大</h5><p>接下来不管是集合合并还是查询，我们都不可避免的需要找到老大，我这里为了方便，直接回传了整个节点的指针。</p>
<p>还是通过递归实现，挺暴力的解法。</p>
<div class="hljs"><pre><code class="hljs cpp"><span class="hljs-function"><span class="hljs-keyword">inline</span> node * <span class="hljs-title">find</span><span class="hljs-params">(node * cur,<span class="hljs-keyword">int</span> x)</span></span>&#123;
    <span class="hljs-keyword">if</span>(cur-&gt;l == cur-&gt;r) <span class="hljs-keyword">return</span> cur; 
    <span class="hljs-keyword">if</span>(x &lt;= cur-&gt;ls-&gt;r) <span class="hljs-keyword">return</span> find(cur-&gt;ls,x);
    <span class="hljs-keyword">if</span>(x &gt;= cur-&gt;rs-&gt;l) <span class="hljs-keyword">return</span> find(cur-&gt;rs,x);
    <span class="hljs-keyword">return</span> <span class="hljs-number">0</span>;
&#125;
<span class="hljs-function"><span class="hljs-keyword">inline</span> node * <span class="hljs-title">get_fa</span><span class="hljs-params">(node * cur,<span class="hljs-keyword">int</span> x)</span></span>&#123;
    node * now = find(cur,x);
    <span class="hljs-keyword">if</span>(x == now-&gt;fa) <span class="hljs-keyword">return</span> now;
    <span class="hljs-keyword">else</span> <span class="hljs-keyword">return</span> get_fa(cur,now-&gt;fa);
&#125;</code></pre></div>
<hr>
<h5 id="集合合并"><a href="#集合合并" class="headerlink" title="集合合并"></a>集合合并</h5><p>首先我们需要找到两个集合的老大，如过现在这两个节点的老大是一样的，说明他们在一个集合里，就不用管他们了(记得版本复制)</p>
<p>如果不同，根据之前的分析，我们就比较他们的深度大小，然后确定谁是新的老大，谁要变成小弟</p>
<div class="hljs"><pre><code class="hljs cpp"> <span class="hljs-keyword">if</span>(opt == <span class="hljs-number">1</span>)&#123;
      <span class="hljs-built_in">cin</span> &gt;&gt; x &gt;&gt; y;
      node * px = get_fa(root[i<span class="hljs-number">-1</span>],x) ,* py = get_fa(root[i<span class="hljs-number">-1</span>],y);
            <span class="hljs-keyword">if</span>(px-&gt;fa == py-&gt;fa)&#123;
            root[i] = root[i<span class="hljs-number">-1</span>];
            <span class="hljs-keyword">continue</span>;
      &#125;
     <span class="hljs-comment">//我们要把 f 的father改成 F </span>
      <span class="hljs-keyword">if</span>(px-&gt;deep &gt; py-&gt;deep) F = px-&gt;fa , f = py-&gt;fa;
      <span class="hljs-keyword">else</span> F = py-&gt;fa , f = px-&gt;fa;
      FLAG = px-&gt;deep == py-&gt;deep;
      root[i] = upd(root[i<span class="hljs-number">-1</span>],f,F);
&#125;</code></pre></div>
<hr>
<h5 id="集合查询"><a href="#集合查询" class="headerlink" title="集合查询"></a>集合查询</h5><p>直接找老大，比较老大</p>
<div class="hljs"><pre><code class="hljs cpp">略</code></pre></div>
<hr>
<h5 id="版本回退"><a href="#版本回退" class="headerlink" title="版本回退"></a>版本回退</h5><p>主席树基操</p>
<div class="hljs"><pre><code class="hljs cpp">略</code></pre></div>
<hr>
<h5 id="版本更新"><a href="#版本更新" class="headerlink" title="版本更新"></a>版本更新</h5><p>这是我卡最久的地方，也是我和题解不一样的地方。</p>
<p><a target="_blank" rel="noopener" href="https://www.luogu.com.cn/blog/88OvO88/solution-p3402">这篇题解</a>里在更新节点深度的时候，直接就更新了上一个版本的深度。确实，这不会对答案的正确性造成影响，但是会一定程度影响效率，个人认为还是有必要要处理的。</p>
<p>所以我们这里要实现的就是，如何在一个版本里修改两个结点。其实很简单，就是在向下传递的时候同时找另外一个点</p>
<div class="hljs"><pre><code class="hljs cpp"><span class="hljs-function"><span class="hljs-keyword">inline</span> node * <span class="hljs-title">upd</span><span class="hljs-params">(node * cur,<span class="hljs-keyword">int</span> x,<span class="hljs-keyword">int</span> F)</span></span>&#123;
    node * now = create();
    copy(now,cur);
    <span class="hljs-keyword">if</span>(cur-&gt;l == cur-&gt;r &amp;&amp; cur-&gt;l == x)&#123;
        now-&gt;fa = F;
        <span class="hljs-keyword">return</span> now;
    &#125; <span class="hljs-keyword">else</span> <span class="hljs-keyword">if</span>(cur-&gt;l == cur-&gt;r &amp;&amp; cur-&gt;l == F)&#123;
        now-&gt;deep = now-&gt;deep + FLAG;
        <span class="hljs-keyword">return</span> now;
    &#125;
    <span class="hljs-keyword">if</span>(cur-&gt;l &lt;= x &amp;&amp; x &lt;= cur-&gt;ls-&gt;r)&#123;
        now-&gt;ls = upd(cur-&gt;ls,x,F);
        <span class="hljs-keyword">if</span>(cur-&gt;rs-&gt;l &lt;= F &amp;&amp; F &lt;= cur-&gt;r &amp;&amp; FLAG) now-&gt;rs = upd(cur-&gt;rs,x,F); <span class="hljs-comment">// 就是这里，同时查找另外一个节点 </span>
    &#125;
    <span class="hljs-keyword">if</span>(cur-&gt;rs-&gt;l &lt;= x &amp;&amp; x &lt;= cur-&gt;r)&#123;
        now-&gt;rs = upd(cur-&gt;rs,x,F);
        <span class="hljs-keyword">if</span>(cur-&gt;l &lt;= F &amp;&amp; F &lt;= cur-&gt;ls-&gt;r &amp;&amp; FLAG) now-&gt;ls = upd(cur-&gt;ls,x,F); <span class="hljs-comment">// 就是这里，同时查找另外一个节点 </span>
    &#125;
    <span class="hljs-keyword">return</span> now;
&#125;</code></pre></div>
<p>其他的也没什么好多说的了，直接放代码吧</p>
<hr>
<h3 id="AC代码"><a href="#AC代码" class="headerlink" title="AC代码"></a>AC代码</h3><div class='spoiler collapsed'>
    <div class='spoiler-title'>
        Code
    </div>
    <div class='spoiler-content'>
        <div class="hljs"><pre><code class="hljs cpp"><span class="hljs-meta">#<span class="hljs-meta-keyword">include</span><span class="hljs-meta-string">&lt;map&gt;</span></span>
<span class="hljs-meta">#<span class="hljs-meta-keyword">include</span><span class="hljs-meta-string">&lt;set&gt;</span></span>
<span class="hljs-meta">#<span class="hljs-meta-keyword">include</span><span class="hljs-meta-string">&lt;cmath&gt;</span></span>
<span class="hljs-meta">#<span class="hljs-meta-keyword">include</span><span class="hljs-meta-string">&lt;queue&gt;</span></span>
<span class="hljs-meta">#<span class="hljs-meta-keyword">include</span><span class="hljs-meta-string">&lt;bitset&gt;</span></span>
<span class="hljs-meta">#<span class="hljs-meta-keyword">include</span><span class="hljs-meta-string">&lt;vector&gt;</span></span>
<span class="hljs-meta">#<span class="hljs-meta-keyword">include</span><span class="hljs-meta-string">&lt;cstdio&gt;</span></span>
<span class="hljs-meta">#<span class="hljs-meta-keyword">include</span><span class="hljs-meta-string">&lt;cstring&gt;</span></span>
<span class="hljs-meta">#<span class="hljs-meta-keyword">include</span><span class="hljs-meta-string">&lt;cstdlib&gt;</span></span>
<span class="hljs-meta">#<span class="hljs-meta-keyword">include</span><span class="hljs-meta-string">&lt;iostream&gt;</span></span>
<span class="hljs-meta">#<span class="hljs-meta-keyword">include</span><span class="hljs-meta-string">&lt;algorithm&gt;</span></span>
<span class="hljs-meta">#<span class="hljs-meta-keyword">define</span> rep(i,a,b) for(register int i = (a);i &lt;= (b);++i)</span>
<span class="hljs-meta">#<span class="hljs-meta-keyword">define</span> per(i,a,b) for(register int i = (a);i &gt;= (b);--i)  </span>
<span class="hljs-keyword">typedef</span> <span class="hljs-keyword">long</span> <span class="hljs-keyword">long</span> ll;
<span class="hljs-keyword">typedef</span> <span class="hljs-keyword">unsigned</span> <span class="hljs-keyword">long</span> <span class="hljs-keyword">long</span> ull;
<span class="hljs-keyword">using</span> <span class="hljs-built_in">std</span>::<span class="hljs-built_in">string</span>;<span class="hljs-keyword">using</span> <span class="hljs-built_in">std</span>::<span class="hljs-built_in">cin</span>;<span class="hljs-keyword">using</span> <span class="hljs-built_in">std</span>::<span class="hljs-built_in">cout</span>;

<span class="hljs-keyword">const</span> <span class="hljs-keyword">int</span> N = <span class="hljs-number">1e5</span>+<span class="hljs-number">10</span>;
<span class="hljs-keyword">int</span> n,m,a[<span class="hljs-number">2</span>*N],opt,tot,FLAG;

<span class="hljs-class"><span class="hljs-keyword">struct</span> <span class="hljs-title">node</span>&#123;</span>
    <span class="hljs-keyword">int</span> l,r,fa,deep,num;
    node * ls, * rs;
&#125;Tree[<span class="hljs-number">32</span>*N],*root[<span class="hljs-number">4</span>*N];

<span class="hljs-function"><span class="hljs-keyword">inline</span> node * <span class="hljs-title">create</span><span class="hljs-params">()</span></span>&#123;<span class="hljs-keyword">return</span> &amp;Tree[++tot];&#125;

<span class="hljs-function"><span class="hljs-keyword">inline</span> <span class="hljs-keyword">void</span> <span class="hljs-title">copy</span><span class="hljs-params">(node * u , node * v)</span></span>&#123;
    u-&gt;l = v-&gt;l , u-&gt;r = v-&gt;r;
    u-&gt;ls = v-&gt;ls , u-&gt;rs = v-&gt;rs;
    u-&gt;fa = v-&gt;fa , u-&gt;num = v-&gt;num , u-&gt;deep = v-&gt;deep;
    <span class="hljs-keyword">return</span>;
&#125;

<span class="hljs-function"><span class="hljs-keyword">inline</span> <span class="hljs-keyword">void</span> <span class="hljs-title">build</span><span class="hljs-params">(node * cur,<span class="hljs-keyword">int</span> l,<span class="hljs-keyword">int</span> r)</span></span>&#123;
    cur-&gt;l = l , cur-&gt;r = r;
    <span class="hljs-keyword">if</span>(l == r)&#123;
        cur-&gt;fa = cur-&gt;num = l , cur-&gt;deep = <span class="hljs-number">1</span>;
        <span class="hljs-keyword">return</span>;
    &#125;
    <span class="hljs-keyword">int</span> mid = (l+r)&gt;&gt;<span class="hljs-number">1</span>;
    cur-&gt;ls = create() , cur-&gt;rs = create();
    build(cur-&gt;ls,l,mid) , build(cur-&gt;rs,mid+<span class="hljs-number">1</span>,r);
    <span class="hljs-keyword">return</span>;
&#125;

<span class="hljs-function"><span class="hljs-keyword">inline</span> node * <span class="hljs-title">upd</span><span class="hljs-params">(node * cur,<span class="hljs-keyword">int</span> x,<span class="hljs-keyword">int</span> F)</span></span>&#123;
    node * now = create();
    copy(now,cur);
    <span class="hljs-keyword">if</span>(cur-&gt;l == cur-&gt;r &amp;&amp; cur-&gt;l == x)&#123;
        now-&gt;fa = F;
        <span class="hljs-keyword">return</span> now;
    &#125; <span class="hljs-keyword">else</span> <span class="hljs-keyword">if</span>(cur-&gt;l == cur-&gt;r &amp;&amp; cur-&gt;l == F)&#123;
        now-&gt;deep = now-&gt;deep + FLAG;
        <span class="hljs-keyword">return</span> now;
    &#125;
    <span class="hljs-keyword">if</span>(cur-&gt;l &lt;= x &amp;&amp; x &lt;= cur-&gt;ls-&gt;r)&#123;
        now-&gt;ls = upd(cur-&gt;ls,x,F);
        <span class="hljs-keyword">if</span>(cur-&gt;rs-&gt;l &lt;= F &amp;&amp; F &lt;= cur-&gt;r &amp;&amp; FLAG) now-&gt;rs = upd(cur-&gt;rs,x,F);
    &#125;
    <span class="hljs-keyword">if</span>(cur-&gt;rs-&gt;l &lt;= x &amp;&amp; x &lt;= cur-&gt;r)&#123;
        now-&gt;rs = upd(cur-&gt;rs,x,F);
        <span class="hljs-keyword">if</span>(cur-&gt;l &lt;= F &amp;&amp; F &lt;= cur-&gt;ls-&gt;r &amp;&amp; FLAG) now-&gt;ls = upd(cur-&gt;ls,x,F);
    &#125;
    <span class="hljs-keyword">return</span> now;
&#125;

<span class="hljs-function"><span class="hljs-keyword">inline</span> node * <span class="hljs-title">find</span><span class="hljs-params">(node * cur,<span class="hljs-keyword">int</span> x)</span></span>&#123;
    <span class="hljs-keyword">if</span>(cur-&gt;l == cur-&gt;r) <span class="hljs-keyword">return</span> cur; 
    <span class="hljs-keyword">if</span>(x &lt;= cur-&gt;ls-&gt;r) <span class="hljs-keyword">return</span> find(cur-&gt;ls,x);
    <span class="hljs-keyword">if</span>(x &gt;= cur-&gt;rs-&gt;l) <span class="hljs-keyword">return</span> find(cur-&gt;rs,x);
    <span class="hljs-keyword">return</span> <span class="hljs-number">0</span>;
&#125;

<span class="hljs-function"><span class="hljs-keyword">inline</span> node * <span class="hljs-title">get_fa</span><span class="hljs-params">(node * cur,<span class="hljs-keyword">int</span> x)</span></span>&#123;
    node * now = find(cur,x);
    <span class="hljs-keyword">if</span>(x == now-&gt;fa) <span class="hljs-keyword">return</span> now;
    <span class="hljs-keyword">else</span> <span class="hljs-keyword">return</span> get_fa(cur,now-&gt;fa);
&#125;

<span class="hljs-function"><span class="hljs-keyword">int</span> <span class="hljs-title">main</span><span class="hljs-params">()</span></span>&#123;
    <span class="hljs-built_in">std</span>::ios::sync_with_stdio(<span class="hljs-number">0</span>);<span class="hljs-built_in">cin</span>.tie(<span class="hljs-number">0</span>);<span class="hljs-built_in">cout</span>.tie(<span class="hljs-number">0</span>);
    <span class="hljs-comment">// freopen(&quot;in.in&quot;,&quot;r&quot;,stdin);</span>
    <span class="hljs-comment">// freopen(&quot;out.out&quot;,&quot;w&quot;,stdout);</span>
    <span class="hljs-keyword">int</span> x,y,F,D,f;
    <span class="hljs-built_in">cin</span> &gt;&gt; n &gt;&gt; m;
    root[<span class="hljs-number">0</span>] = create();
    build(root[<span class="hljs-number">0</span>],<span class="hljs-number">1</span>,n);
    rep(i,<span class="hljs-number">1</span>,m)&#123;
        <span class="hljs-built_in">cin</span> &gt;&gt; opt;
        <span class="hljs-keyword">if</span>(opt == <span class="hljs-number">1</span>)&#123;
            <span class="hljs-built_in">cin</span> &gt;&gt; x &gt;&gt; y;
            node * px = get_fa(root[i<span class="hljs-number">-1</span>],x) ,* py = get_fa(root[i<span class="hljs-number">-1</span>],y);
            <span class="hljs-keyword">if</span>(px-&gt;fa == py-&gt;fa)&#123;
                root[i] = root[i<span class="hljs-number">-1</span>];
                <span class="hljs-keyword">continue</span>;
            &#125;
            <span class="hljs-keyword">if</span>(px-&gt;deep &gt; py-&gt;deep) F = px-&gt;fa , f = py-&gt;fa;
            <span class="hljs-keyword">else</span> F = py-&gt;fa , f = px-&gt;fa;
            FLAG = px-&gt;deep == py-&gt;deep;
            root[i] = upd(root[i<span class="hljs-number">-1</span>],f,F);
        &#125; <span class="hljs-keyword">else</span> <span class="hljs-keyword">if</span>(opt == <span class="hljs-number">2</span>)&#123;
            <span class="hljs-built_in">cin</span> &gt;&gt; x;
            root[i] = root[x];
        &#125; <span class="hljs-keyword">else</span> &#123;
            root[i] = root[i<span class="hljs-number">-1</span>];
            <span class="hljs-built_in">cin</span> &gt;&gt; x &gt;&gt; y;
            node * px = get_fa(root[i],x) ,* py = get_fa(root[i],y);
            <span class="hljs-keyword">if</span>(px-&gt;fa == py-&gt;fa) <span class="hljs-built_in">cout</span> &lt;&lt; <span class="hljs-string">&quot;1\n&quot;</span>;
            <span class="hljs-keyword">else</span> <span class="hljs-built_in">cout</span> &lt;&lt; <span class="hljs-string">&quot;0\n&quot;</span>;
        &#125;
    &#125;
    <span class="hljs-keyword">return</span> <span class="hljs-number">0</span>;
&#125;</code></pre></div>

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            "font-weight": "bold",
            "color": rand_color()
          });
          // 随机颜色
          function rand_color() {
            return "rgb(" + ~~(255 * Math.random()) + "," + ~~(255 * Math.random()) + "," + ~~(255 * Math.random()) + ")"
          }
          //在body添加这个标签
          $("body").append($i);
          //animate() 方法执行 CSS 属性集的自定义动画。
          //该方法通过CSS样式将元素从一个状态改变为另一个状态。CSS属性值是逐渐改变的，这样就可以创建动画效果。
          //详情请看http://www.w3school.com.cn/jquery/effect_animate.asp
          $i.animate({
            //将原来的位置向上移动180
            "top": y - 180,
            "opacity": 0
            //1500动画的速度
          }, 1500, function () {
            //时间到了自动删除
            $i.remove();
          });
        });
      })
      ;
    </script>
  














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